Simplify; express your answer in exponential form. Assume $t\neq 0, r\neq 0$. $\dfrac{{(t^{3}r)^{5}}}{{(t^{-1}r^{2})^{-4}}}$
Solution: To start, try simplifying the numerator and the denominator independently. In the numerator, we can use the distributive property of exponents. ${(t^{3}r)^{5} = (t^{3})^{5}(r)^{5}}$ On the left, we have ${t^{3}}$ to the exponent ${5}$ . Now ${3 \times 5 = 15}$ , so ${(t^{3})^{5} = t^{15}}$ Apply the ideas above to simplify the equation. $\dfrac{{(t^{3}r)^{5}}}{{(t^{-1}r^{2})^{-4}}} = \dfrac{{t^{15}r^{5}}}{{t^{4}r^{-8}}}$ Break up the equation by variable and simplify. $\dfrac{{t^{15}r^{5}}}{{t^{4}r^{-8}}} = \dfrac{{t^{15}}}{{t^{4}}} \cdot \dfrac{{r^{5}}}{{r^{-8}}} = t^{{15} - {4}} \cdot r^{{5} - {(-8)}} = t^{11}r^{13}$